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Ahmed1029

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In summary, Faraday's law of induction, in its differential form, states that the curl of the electric field is equal to the negative time derivative of the magnetic field. The integral form of the law, obtained through Stokes's integral theorem, shows that the line integral of the electric field is equal to the negative time derivative of the magnetic flux through an oriented surface. However, this integral form can only be used in situations with certain symmetries and requires prior knowledge of the direction of the electric field in order to solve for it.

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Ahmed1029

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BvU

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Ahmed1029

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Lenz's law determines the direction of the current. Is the induced field exactly in the direction of the current?BvU said:

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Hint:Ahmed1029 said:Lenz's law determines the direction of the current. Is the induced field exactly in the direction of the current?

-Dan

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hutchphd

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Or perhaps more directly $$\mathbf J =\sigma \mathbf E $$ using current density and conductivity

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Ahmed1029

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I know within the current-occupied volume it has to be this way, but what about other points?

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What is the direction of the field at other points? Are you asking about what happensAhmed1029 said:I know within the current-occupied volume it has to be this way, but what about other points?

-Dan

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Ahmed1029

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No, I meant there is a changing magnetic field outside the interior of the cable itself which would induce a curly electric field. But I realized it was unnecessary to know this piece of detail; Your previous answers suffice.topsquark said:What is the direction of the field at other points? Are you asking about what happensoutsidethe region of flux change?

-Dan

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$$\vec{\nabla} \times \vec{E} = -\partial_t \vec{B}.$$

You get the integral law by using Stokes's integral theorem, i.e., you integrate the equation over an oriented surface, ##A##, i.e., a surface with surface-normal vectors ##\mathrm{d}^2 \vec{f}## oriented arbitrarily in one of two possible directions. Then on the left-hand side you can transform the surface integral as a line integral along the boundary, ##\partial A##, of the surface. Its orientation must be chosen using the right-hand-rule relative to the orientation of the surface-normal vectors. Then you get

$$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E} = -\int_A \mathrm{d}^2 \vec{f} \cdot \partial_t \vec{B}.$$

Finally, if we assume for simplicity that the surface and its boundary are at rest (and ONLY then) you can take the time derivative out of the surface integral

$$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E} = -\frac{\mathrm{d}}{\mathrm{d} t} \int_A \mathrm{d}^2 \vec{f} \cdot \vec{B}.$$

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Ahmed1029

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But you will have to know the direction of the electric field beforehand in order to apply that last equation in a way that gets you the electric field, which is what I'm asking.vanhees71 said:

$$\vec{\nabla} \times \vec{E} = -\partial_t \vec{B}.$$

You get the integral law by using Stokes's integral theorem, i.e., you integrate the equation over an oriented surface, ##A##, i.e., a surface with surface-normal vectors ##\mathrm{d}^2 \vec{f}## oriented arbitrarily in one of two possible directions. Then on the left-hand side you can transform the surface integral as a line integral along the boundary, ##\partial A##, of the surface. Its orientation must be chosen using the right-hand-rule relative to the orientation of the surface-normal vectors. Then you get

$$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E} = -\int_A \mathrm{d}^2 \vec{f} \cdot \partial_t \vec{B}.$$

Finally, if we assume for simplicity that the surface and its boundary are at rest (and ONLY then) you can take the time derivative out of the surface integral

$$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E} = -\frac{\mathrm{d}}{\mathrm{d} t} \int_A \mathrm{d}^2 \vec{f} \cdot \vec{B}.$$

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The direction of an induced electric field is determined by the right-hand rule. This rule states that if you point your right thumb in the direction of the changing magnetic field, then the curled fingers will point in the direction of the induced electric field.

Yes, the direction of an induced electric field can change. It is dependent on the direction and strength of the changing magnetic field.

The direction of an induced electric field is affected by the strength and direction of the changing magnetic field, as well as the material through which the field is passing.

The direction of an induced electric field in a coil can be determined using the right-hand rule. If the coil is wrapped around a core, the direction of the induced electric field will be determined by the direction of the changing magnetic field passing through the core.

No, the direction of an induced electric field is not always opposite to the direction of the changing magnetic field. It can be in the same direction or perpendicular to the changing magnetic field, depending on the specific circumstances.

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